If $G$ is a Lie group and we take its loop group $LG$ why do we deal with projective representations of $LG$ and central extensions thereof? Where does the extra complexity come in to require us to consider this extra, very complicated, step?

When are you ever required to do anything in mathematics?
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Qiaochu YuanJul 24 '12 at 1:56

I asked this because I am reading about these on behest of my advisor with little motivation. I told him I was keen on mathematical physics and he told me to read Loop groups by Segal and Pressley.
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SvenJul 25 '12 at 1:53

It just seems like strange wording to me. Nobody is required to centrally extend a loop group. People want to centrally extend loop groups.
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Qiaochu YuanJul 25 '12 at 2:13

2 Answers
2

Projective representations are needed in quantum mechanics, and I think that is the main motivation behind studying projective representations. The point is that in QM, it is rays in a Hilbert space which have physical meaning. So the symmetry group is required to act on and transform these rays rather than vectors.

So the representations we study happen to be projective because of some physical meaning? Is there no mathematical reason?
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SvenJul 23 '12 at 22:29

I am not sure. If there is any mathematical reason I too would like to know :)
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user10001Jul 23 '12 at 22:41

But from Lie algebra point of view study of projective representations will also give you ordinary ones. Right ?
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user10001Jul 23 '12 at 22:44

@Sven: I disagree with the premise that a physical justification for studying projective representations is not a mathematical justification for studying projective representations. Much of mathematics owes its existence to physical considerations and this seems disrespectful of the role of physics in mathematics to me.
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Qiaochu YuanJul 24 '12 at 1:55

I meant no offense, if it were a physical constraint I would be perfectly happy with that explanation. I asked because in my limited experience the physical constraint is not independent of a mathematical one.
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SvenJul 25 '12 at 1:48

Well, we are usually concerned about these exotic objects at the algebraic level. They pop up when we consider spinor representations of the orthogonal group. Lets write $\gamma_{\alpha}$ for the generators of the Clifford algebra which are represented by $\Gamma_{\alpha}$. They satisfy
$$\Gamma_{\alpha}\Gamma_{\beta}+\Gamma_{\beta}\Gamma_{\alpha}=2\eta_{\alpha\beta}$$
where $\eta_{\alpha\beta}$ is nondegenerate. (If it's degenerate, we have a Grassmann algebra and not a Clifford algebra, but it's not terrible.)

So we have commutators of the generators of the Clifford algebra. We will leave it as an exercise for the reader to determine
$$[\Gamma_{\alpha},\Gamma_{\beta}]=\text{something quadratic}$$
and
$$[\Gamma_{\alpha},\Gamma_{\mu}\Gamma_{\nu}]=2\eta_{\alpha\mu}\Gamma_{\nu}-2\eta_{\alpha\nu}\Gamma_{\mu}.$$
Now we expect commutators of quadratic terms to look like
$$[\Gamma_{\alpha}\Gamma_{\beta},\Gamma_{\mu}\Gamma_{\nu}]\sim a\Gamma_{\sigma}\Gamma_{\tau}+b\cdot\textbf{1}$$
and wouldn't it be lovely if we could make this an equality and ditch the constant multiple of the identity operator $b\textbf{1}$? This would make the representation of quadratic guys a closed algebra!

We find if we write an arbitrary element of these generated algebra as $a\cdot\textbf{1}+b^{\mu\nu}\Gamma_{\mu}\Gamma_{\nu}$ (where we use Einstein summation convention) we have an algebra, provided one condition on the matrix $b^{\mu\nu}$ namely we need
$$ b^{\mu\nu}=-b^{\nu\mu}.$$
This says we have a "representation" of the orthogonal Lie algebra. Is it a representation? Yes and no. No, it's technically not a "vanilla representation"; but with all this extra structure, we have a central extension of the orthogonal lie algebra.

Wouldn't it be lovely, after all this work, if we could do the same thing to other Lie algebras? That's how we get started with losing ourselves in the jungle of loop groups, its central extensions, etc.